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#1
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There is an event that happens 1/10 times. It costs $1 for each attempt. If the attempt succeeds, you get $9 back, for a profit of $8. If you play up to 10 times, but stop when you win, is it profitable?
I know this shouldn't be profitable, but let me explain some logic and tell me why it's faulty? In a sequence of 10 events, NOT STOPPING, the event will be successful once. In this sequence, it will evenly be distributed throughout each try, averaging at try #5.5. Visual Display: X = hit 10 trials of 10 sequences X _ _ _ _ _ _ _ _ _ _ X _ _ _ _ _ _ _ _ _ _ X _ _ _ _ _ _ _ _ _ _ X _ _ _ _ _ _ _ _ _ _ X _ _ _ _ _ _ _ _ _ _ X _ _ _ _ _ _ _ _ _ _ X _ _ _ _ _ _ _ _ _ _ X _ _ _ _ _ _ _ _ _ _ X _ _ _ _ _ _ _ _ _ _ X This is average, yes? So if you go through one sequence, stopping when you hit, it should be profitable? |
#2
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Ok, to clarify more:
In a sequence of 10 tries, you will succeed 1 time, on average. If you have infinite sequences of 10 tries, 1/10 times, the success will be on the first try. 1/10 times, the success will be on the second try, etc. |
#3
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My calculations:
Note - You STOP at the X. Visual Display: X = hit 10 trials of 10 sequences X _ _ _ _ _ _ _ _ _ +8 _ X _ _ _ _ _ _ _ _ +7 _ _ X _ _ _ _ _ _ _ +6 _ _ _ X _ _ _ _ _ _ +5 _ _ _ _ X _ _ _ _ _ +4 _ _ _ _ _ X _ _ _ _ +3 _ _ _ _ _ _ X _ _ _ +2 _ _ _ _ _ _ _ X _ _ +1 _ _ _ _ _ _ _ _ X _ 0 _ _ _ _ _ _ _ _ _ X -1 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 + 0 - 1 = 35 35/10 = 3.5 |
#4
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you're assuming that you will automatically get at least one in ten X, implying that the probability of each trial is not independent. if the probability of each draw is 1/10 than it would be a geometric distribution and the average number of draws before a hit would be 10. meaning a loss of $.10 on each trial.
if a miss implies a greater likelyhood of a hit later on than the probability would be completely differnt, i could do it out but i don't think that's what you were wondering about, if it is than just respond and i'll show you the calculations -little fishy |
#5
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sorry i responded before i saw your update... i'll do the math out again tomorrow, and if you succeeed exactly once out of every ten trials and yuou can stop after you suceed then yes a 1:8 wager would be EV, more on this tomorrow, but tonight it's bed time
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#6
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Your EV =
1st trial = (0.1) * $8 = $0.8 2nd trial = (0.9)(0.1) $7 = $0.63 3rd trial = (0.9)^2(0.1) $6 = $0.486 4th trial = (0.9)^3(0.1) $5 = $0.3645 5th trial = (0.9)^4(0.1) $4 = $0.26244 6th trial = (0.9)^5(0.1) $3 = $0.177147 7th trial = (0.9)^6(0.1) $2 = $0.1062882 8th trial = (0.9)^7(0.1) $1 = $0.04782969 9th trial = (0.9)^8(0.1) $0 = doesnt matter 10th trial= (0.9)^9(0.1) -$1 = -$0.03874205 Dont win at all = (0.9)^10 * -$10 = -$3.486784 E(X) = -$0.65 P.S. Theres a 57% chance you will win by the 8th trial and be therefore profitable. |
#8
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I know the game is -EV. I want to know why my line of thought is wrong.
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#9
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I wasn't replying to you, I was replying to someone who claimed the game is +EV. I don't understand what you mean yet.
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#10
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[ QUOTE ]
My calculations: Note - You STOP at the X. Visual Display: X = hit 10 trials of 10 sequences X _ _ _ _ _ _ _ _ _ +8 _ X _ _ _ _ _ _ _ _ +7 _ _ X _ _ _ _ _ _ _ +6 _ _ _ X _ _ _ _ _ _ +5 _ _ _ _ X _ _ _ _ _ +4 _ _ _ _ _ X _ _ _ _ +3 _ _ _ _ _ _ X _ _ _ +2 _ _ _ _ _ _ _ X _ _ +1 _ _ _ _ _ _ _ _ X _ 0 _ _ _ _ _ _ _ _ _ X -1 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 + 0 - 1 = 35 35/10 = 3.5 [/ QUOTE ] This did not weight the likelihood of the events correctly. |
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